\(2014.2013.2013.2012.x=2011.2012.2013.2014\)

\(\Rightarrow x=\dfrac{2011.2012.2013.2014}{2014.2013.2013.2012}\)

\(\Rightarrow x=\dfrac{2011}{2013}\)

Bạn xem lại đề

1/

Số cần tìm \(\overline{ab7}\) theo đề bài

\(\overline{7ab}=2.\overline{ab7}+21\)

\(\Rightarrow700+\overline{ab}=20.\overline{ab}+14+21\)

\(\Leftrightarrow19.\overline{ab}=665\Rightarrow\overline{ab}=665:19=35\)

Số cần tìm là 357

2/

Gọi số cần tìm là \(\overline{ab}\) theo đề bài

\(\overline{ba}-\overline{ab}=63\)

\(10.b+a-10.a-b=63\)

\(9.\left(b-a\right)=63\Rightarrow b-a=7\)

\(a=\left(9-7\right):2=1\)

\(\Rightarrow b=9-a=9-1=8\)

Số cần tìm là 18

a) \(2x^2-3xy-2y^2=2\)

\(\Rightarrow2x^2+xy-4xy-2y^2=2\)

\(\Rightarrow x\left(2x+y\right)-2y\left(2x+y\right)=2\)

\(\Rightarrow\left(2x+y\right)\left(x-2y\right)=2\)

\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)

Ta giải các hệ phương trình sau với x;y nguyên 

1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

4)  \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)

b) \(xy-y+x=9\)

\(\Rightarrow y\left(x-1\right)+x-1+1=9\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)

A = 235\(\times\) 106 - 24255 : [ 240 - 9]

A = 24910  - 105

A = 24805

2²ˣ⁻³ = 32

2²ˣ⁻³ = 2⁵

2x - 3 = 5

2x = 5 + 3

2x = 8

x = 8 : 2

x = 4

x = 4

\(50\%+\dfrac{1}{2}+\dfrac{1}{7}x\dfrac{2}{5}+\dfrac{1}{7}x\dfrac{3}{5}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{7}x\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=1+\dfrac{1}{7}x1\)

\(=\dfrac{7}{7}+\dfrac{1}{7}=\dfrac{8}{7}\)

50% + 1/2 + 1/7 × 2/5 + 1/7 × 3/5

= 1/2 + 1/2 + 1/7 × (2/5 + 3/5)

= 1 + 1/7 × 1

= 1 + 1/7

= 7/7 + 1/7

= 8/7

\(\overline{1xy2z}\) \(⋮\) 1125

1125 = 3².5³ ⇒ \(\overline{1xy2z}\) ⋮ 9; 125

\(\overline{1xy2z}\) ⋮ 125 ⇒\(\overline{...00}\);  \(\overline{..25}\);  \(\overline{...50}\);  \(\overline{..75}\) ⇒ \(z\) = 5

⇒ \(\overline{1xy25}\) ⋮ 125 ;    \(\overline{1xy25}\)  = \(\overline{1x}\) \(\times\) 1000 + \(\overline{y25}\) ⋮ 125

⇒ \(\overline{y25}\) ⋮ 125 ⇒ \(\overline{y25}\) \(\in\) { 125; 250; 375; 500; 625; 750; 875; 1000; ..;}

⇒ \(\overline{y25}\) = 125; 625 ⇒ y = 1; 6 

vì \(\overline{1xy25}\) ⋮ 9 ⇒ 1 + \(x\) + y + 2 + 5 ⋮ 9 ⇒ \(x+y\) + 8 ⋮ 9 (1)

Thay y = 1 vào  (1) ta có: \(x\) + 1 + 8 ⋮ 9 ⇒ \(x\) = 0; 9 

\(\overline{1xy25}\) = 10125;  19125

Thay y = 6 vào (1) ta có: \(x+6+8\) ⋮ 9 ⇒ \(x\) = 4

⇒ \(\overline{1xy25}\) = \(14625\)

Kết luận các số thỏa mãn đề bài lần lượt là:

10125; 14625; 19125